求lim(n→+∞ ) (1+n)[ln(1+n)-ln n]

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求lim(n→+∞ ) (1+n)[ln(1+n)-ln n]

求lim(n→+∞ ) (1+n)[ln(1+n)-ln n]
求lim(n→+∞ ) (1+n)[ln(1+n)-ln n]

求lim(n→+∞ ) (1+n)[ln(1+n)-ln n]

答案是1 原式=lim(n→+∞ ) [ln(1+n)-ln n]/[1/(1+n)] --洛必达->
lim(n→+∞ )(n+1)/n=1

lim(n→+∞ ) (1+n)[ln(1+n)-ln n]
=lim(n→+∞ ) (1+n)[ln(1+n)/n]
=lim(n→+∞ ) (1+n)[ln(1+1/n)]
=lim(n→+∞ ) [ln(1+1/n)^(n+1)]
=lim(n→+∞ ) [ln(1+1/n)^n*(1+1/n)]
=ln[lim(n→+∞ )(1+1/n)^n*lim(n→+∞ )(1+1/n)]
=ln(e*1)
=1